Internal Combustion Engine MCQ Quiz - Objective Question with Answer for Internal Combustion Engine - Download Free PDF

Explanation:

Internal Combustion Engines (IC Engines):

• Internal Combustion (IC) engines are mechanical devices that convert chemical energy from fuel into mechanical energy by burning the fuel-air mixture inside a combustion chamber. These engines are widely used in automobiles, power plants, and other industrial applications. Depending on various parameters, IC engines are classified into different categories. One such classification is based on the method of fuel injection, which divides IC engines into carburetor engines and air injection engines.

Method of Fuel Injection

This classification is based on how the fuel is supplied to the combustion chamber:

• Carburetor Engine: In this type of engine, a carburetor is used to mix air and fuel in the correct proportion before it enters the combustion chamber. The carburetor plays a crucial role in maintaining the air-fuel ratio, ensuring proper combustion. These engines are commonly found in older vehicles and smaller engines, such as motorcycles and lawnmowers. However, carburetor engines have been largely replaced by more advanced fuel injection systems in modern vehicles due to their limitations in efficiency and precision.
• Air Injection Engine: In air injection engines, fuel is injected directly into the combustion chamber using a high-pressure air stream. This method ensures better atomization of fuel, leading to improved combustion efficiency and reduced emissions. Air injection engines are often used in heavy-duty applications, such as large diesel engines, where precise fuel delivery and combustion control are crucial.

Explanation:

Two-Stroke Petrol Engine:

• A two-stroke petrol engine is a type of internal combustion engine that completes a power cycle in two strokes of the piston during only one crankshaft revolution. In this type of engine, the inlet port, transfer port, and exhaust port play crucial roles in ensuring the proper intake of the air-fuel mixture, transfer of the mixture to the combustion chamber, and expulsion of exhaust gases.
• The opening and closing of the inlet port in a two-stroke engine are controlled by the movement of the piston itself. Unlike in four-stroke engines, there are no valves in a traditional two-stroke engine; instead, ports are used. The timing of these ports is critical for the efficient operation of the engine.

30° to 40° before BDC:

• The inlet port begins to open when the piston is moving towards the bottom dead center (BDC), which allows the fresh air-fuel mixture to enter the crankcase. This occurs slightly before the piston reaches BDC, typically in the range of 30° to 40° before BDC, to ensure that the mixture has enough time to flow into the crankcase under the influence of the pressure difference created by the piston's motion.

Working Mechanism:

1. Induction Phase: As the piston moves downward during its power stroke, it simultaneously uncovers the inlet port. This movement creates a low-pressure area in the crankcase. The fresh air-fuel mixture is then drawn into the crankcase through the inlet port due to the pressure difference.

2. Timing of Inlet Port Opening: The inlet port opens slightly before the piston reaches BDC, as specified in the range of 30° to 40° before BDC. This ensures that the air-fuel mixture starts entering the crankcase at the optimal time, maximizing the engine's efficiency and power output.

3. Transfer and Exhaust Phases: Once the piston starts moving upward after reaching BDC, it compresses the air-fuel mixture in the crankcase. At the same time, the transfer port and exhaust port open to allow the fresh mixture to enter the combustion chamber and the exhaust gases to exit, respectively.

Explanation:

Mean Effective Pressure (MEP) in Engine Performance Analysis

• The Mean Effective Pressure (MEP) is a critical parameter in engine performance analysis. It represents the average pressure acting on the piston during the entire engine cycle, which would produce the measured work output if it were applied uniformly. MEP is not an actual physical pressure but a theoretical concept used to evaluate the performance and efficiency of internal combustion engines.
• In simpler terms, MEP provides a way to compare the work output of engines irrespective of their size, speed, or displacement. This parameter is especially useful for engineers and designers to assess and optimize engine designs.

\(mep = \frac{{{W_{net}}}}{{{V_s}}}\)

For Otto cycle:

\({W_{net}} = {Q_1} - {Q_2} = m{C_v}\left[ {\left( {{T_3} - {T_2}} \right) - \left( {{T_4} - {T_1}} \right)} \right]\)

\({V_s} = {V_1} - {V_2} = {V_1}\left( {1 - \frac{{{V_2}}}{{{V_1}}}} \right) = {V_1}\left( {1 - \frac{1}{r}} \right)\)

\({V_s} = {V_1}\left( {\frac{{r - 1}}{r}} \right) = \frac{{mR{T_1}}}{{{P_1}}}\left( {\frac{{r - 1}}{r}} \right) = \frac{{m{C_v}\left( {r - 1} \right){T_1}}}{{{P_1}}}\left( {\frac{{r - 1}}{r}} \right)\)

\(mep = \frac{{{W_{net}}}}{{{V_s}}} = \frac{{m{C_v}\left[ {\left( {{T_3} - {T_2}} \right) - \left( {{T_4} - {T_1}} \right)} \right]}}{{\frac{{m{C_v}\left( {\gamma - 1} \right){T_1}}}{{{P_1}}}\left( {\frac{{r - 1}}{r}} \right)}}\)

Explanation:

Simple Carburetor:

• A simple carburetor is a critical component in petrol (spark-ignition) engines. Its primary function is to atomize and mix fuel with air in the proper ratio required for efficient combustion. The carburetor ensures that the engine receives the correct air-fuel mixture for various operating conditions, such as idling, acceleration, cruising, and deceleration.
• In a petrol engine, the combustion process requires a specific air-fuel ratio for optimal performance. The ideal ratio, known as the stoichiometric ratio, is approximately 14.7:1, meaning 14.7 parts of air to 1 part of fuel by weight. However, depending on the operating conditions, the engine may require a richer (more fuel) or leaner (less fuel) mixture. The carburetor is designed to adjust the air-fuel mixture dynamically to meet these requirements.

Working of a Simple Carburetor:

• Air Flow: Air enters the carburetor through the air intake. The airflow is controlled by a throttle valve, which adjusts the engine's power output by changing the volume of air entering the engine.
• Fuel Atomization: Fuel is drawn from the fuel reservoir into the carburetor's venturi, a narrow passage where the air velocity increases, creating a low-pressure zone. This pressure difference causes the fuel to atomize, breaking it into tiny droplets.
• Air-Fuel Mixing: The atomized fuel mixes with the incoming air to form a homogenous air-fuel mixture. The mixture's composition is regulated using jets and needles within the carburetor.
• Delivery to Combustion Chamber: The air-fuel mixture is then delivered to the engine's combustion chamber, where it is ignited by the spark plug to produce power.

Explanation:

Requirements of an Ignition System in a Spark-Ignition Engine:

An ignition system is a critical component of a spark-ignition engine, designed to ignite the air-fuel mixture in the combustion chamber at the right time to ensure efficient combustion and power generation. The ignition system must meet several requirements to function effectively in a spark-ignition engine. These requirements include:

1. Timing the Spark: The ignition system must precisely time the spark to coincide with the compression stroke, ensuring that the air-fuel mixture is ignited when it is optimally compressed. Proper timing is crucial for maximizing the engine's power output and efficiency.
2. Generating High Voltage: The ignition system must generate sufficient voltage to jump the gap between the electrodes of the spark plug, creating a spark that can ignite the air-fuel mixture.
3. Maintaining Consistent Spark Duration: The ignition system must maintain a consistent spark duration across all engine speeds (RPMs) to ensure reliable ignition under varying operating conditions.
4. Reliability and Durability: The ignition system must be reliable and durable to withstand the harsh conditions inside an engine, including high temperatures, vibrations, and exposure to combustion by-products.

Mixing air and fuel in the intake manifold.

• This is NOT a requirement of the ignition system in a spark-ignition engine because the function of mixing air and fuel is performed by the carburetor (in older engines) or the fuel injection system (in modern engines). These components are responsible for preparing the air-fuel mixture and delivering it to the combustion chamber. The ignition system's role begins after the air-fuel mixture has already been prepared and compressed in the cylinder. Its primary function is to generate a spark to ignite the mixture, not to mix it.

Concept:

Diesel cycle:

\(Comression\;ratio\left( r \right) = \frac{{{v_1}}}{{{v_2}}}\)

\(Cutt - off\;ratio\left( {{ρ}} \right) = \frac{{{v_3}}}{{{v_2}}}\)

If cut-off happens at k % of the stroke, then

cut-off ratio (ρ) = 1 + k(r - 1)

Calculation:

Given:

r = 16, k = 8 % , ρ = ?

(ρ) = 1 + k(r - 1) 

∴ 1 + 0.08 (16 - 1) = 2.20

Concept:

• Whenever the engine is started from cold, the coolant temperature has to be brought to the desired level in order to minimize the warm-up time.
• This purpose is achieved by a thermostat fitted in a system which initially prevents the circulation of water below a certain temperature through the radiator so that the water gets heated up quickly.
• When the preset temperature is reached, the thermostat allows the water to flow through the radiator.

Concept:

Brake specific fuel consumption (BSFC) =m_f/BP

Where m_f = mass flow rate of fuel, BP = Brake Power

\(Brake\;thermal\;efficiency\left( {{\eta _b}} \right) = \frac{{BP}}{{{m_f} \times CV}} = \frac{1}{{BSFC \times CV}}\)

CV = Calorific Value

Calculation:

Given:

CV = 40 MJ/kg = 40 × 10⁶ J/kg.

\(BSFC = 200\;gm/kWh = \frac{{200 \times {{10}^{ - 3}}}}{{\left( {3600 \times {{10}^3}} \right)}}\;kg/J = \frac{1}{{18}} \times {10^{ - 6}}\;kg/J\)

\(\eta = \frac{1}{{\left( {\frac{1}{{18}}} \right) \times {{10}^{ - 6}} \times 40 \times {{10}^6}}} = \frac{{18}}{{40}} = 0.45 = 45\% \)

Concept:

The capacity of engine is given by:

Capacity of engine = Swept volume × Numbers of cylinders(n)

Swept volume is given by:

\(Swept ~volume= \frac{\pi }{4} \times {d^2} \times l\)

Calculation:

Given:

d = 4 cm, L = 7 cm, n = 4

Clearance volume, V_{c ­}= 2 cm³

Capacity of engine is:

capacity of engine = Swept volume × Numbers of cylinders 

\(Capacity~of~engine = \frac{\pi }{4} \times {d^2} \times l \times n = \frac{\pi }{4} \times {4^2} \times 7 \times 4 = 352~cm^3\)

Concept:

Diesel cycle:

Processes in compression engine (diesel cycle) are:

Process 1-2: Reversible adiabatic compression

Process 2-3: Constant pressure heat addition

Process 3-4: Reversible adiabatic expansion

Process 4-1: Constant volume of heat rejection

cut-off ratio:

The cut-off ratio is the ratio of the volume after combustion to the volume before combustion.

Cut-off ratio: \({r_c} = \frac{{{V_3}}}{{{V_2}}}\)

Compression ratio: \({r} = \frac{{{V_1}}}{{{V_2}}} \)

The efficiency of the diesel cycle is given by

\(\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}\left[ {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right]\)

The mean effective pressure (p_m) which is an indication of the internal work output increases with a pressure ratio at a fixed value of compression ratio and the ratio of specific heats.

The expression for mean effective pressure for diesel cycle,

\({p_m} = \frac{{{p_1}\left[ {\gamma {r^\gamma }\left( {{r_c} - 1} \right) - r\left( {r_c^\gamma - 1} \right)} \right]}}{{\left( {\gamma - 1} \right)\left( {r - 1} \right)}}\)

From the expression,

The mean effective pressure of the diesel cycle having a fixed compression ratio will increase if the cut-off ratio increases.

Explanation:

There are three standard cycles that are used to perform analysis of IC engine:
1) Constant volume combustion (Otto) cycle

2) Constant pressure combustion (Diesel) cycle

3) Combination of constant volume and constant pressure combustion (Dual) cycle

Assumptions during analysis:

• The working fluid throughout the cycle is air and it is treated as an ideal gas 
• The compression and expansion processes are taken as frictionless and adiabatic (no heat loss) i.e. they are reversible 
• The chemical equilibrium of the working fluid is taken as constant   
• The combustion process is replaced by well-defined heat addition processes
• The exhaust process is replaced by a heat rejection process that returns the air of the cycle to intake conditions
• Since the gas is assumed as ideal the specific heats at constant volume and pressure are taken as constant 

​∴ There are no intake and exhaust processes because they are replaced by heat addition and heat rejection processes

Concept:

Diesel cycle:

P-V and T-S diagram of Diesel cycle are:

Compression ratio (r) is given by:

 \(r = \frac{{{v_1}}}{{{v_2}}}\)

Cut-off ratio (r_c) is given by:

\( {{r_c}} = \frac{{{v_3}}}{{{v_2}}}\)

Calculation:

Given:

Compression ratio (r) = 16 = \(\frac{{{v_1}}}{{{v_2}}}\)

v₃ - v₂ = 0.06(v₁ - v₂)

\(\frac{{{v_3}}}{{{v_2}}} - 1 = \frac{6}{{100}}\left( {\frac{{{v_1}}}{{{v_2}}} - 1} \right)\)

\({r_c} - 1 = \frac{6}{{100}}\left( {r - 1} \right)\)

\({r_c} - 1 = \frac{6}{{100}}\left( {16 - 1} \right)\)

r_c = 1.9

Concept:

Thermal efficiency of Otto Cycle:

\({\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\)

Compression ratio: r = v₁/v₂

\(\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{\gamma \; - \;1}}{\gamma }}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma \; - \;1}}\)

\(\frac{{{V_1}}}{{{V_2}}} = r\)

\(\frac{{{T_2}}}{{{T_1}}} = {\left( r \right)^{\gamma - 1}}\)

\({\eta _{otto}} = 1 - \frac{1}{{{{\left( r \right)}^{\gamma - 1}}}}\; = 1 - \frac{1}{{\frac{{{T_2}}}{{{T_1}}}}} = 1 - \frac{{{T_1}}}{{{T_2}}}\)

It is given that \({\eta _{otto}} = 1 - \frac{{{T_a}}}{{{T_b}}}\)

Comparing it to the derived equation, T_a resembles T₁ and T_b resembles T_2. 

T₂ is the temperature where compression stops and the constant volume heat addition starts.

∴ T_b is the temperature where constant volume heat addition starts.

Concept:

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Calculation:

Given: 

Brake power (BP) = 200 kW, Half load = 100 kW Friction Power (FP) = 25 kW

Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)

Mechanical efficiency at half load \( = \frac{{100}}{{125 }}\)

Mechanical efficiency at half load = 0.8 ⇒ 80 %

Explanation:

Petroil (Petro-oil lubrication system): In this method, the lubricating oil is mixed with petrol and fed into the engine cylinder during the suction stroke. The droplets of the partials cause the lubricating effect in the engine cylinder.

This method of lubrication is used in small engines like motorcycles and scooters. The system of lubrication is used in scooters and motorcycles, particularly for two-stroke engines about 3 to 6% of lubrication oil is added with petrol is the petrol tank.

The petrol evaporates when the engine is working. The lubricating oil is left behind in the form of mist. The parts of the engine such as piston, cylinder walls and connecting rod are lubricated by being waited with the oil mist left behind.

Splash lubrication system: The splashing action of oil maintains a fog or mist of oil that drenches the inner parts of the engine such as bearings, cylinder walls, pistons, piston pins, timing gears etc. The splash oil then drips back into the sump.

This system is commonly used in a single-cylinder engine with the closed crankcase.